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Write phosphorous reacts with aqueous caustic soda to give hypophosphite and phosphine. |
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Answer» Solution :a) The ionic skeleton equation is `P_(4)overset(OH^(-))rarrPH_(3)+H_(2)PO_(2)^(-)` b) Writing oxidation numbers `overset(0)(P_(4))rarroverset(-3)(P)overset(+1)(H_(3))+overset(+1)(H_(2))overset(+1)(H_(2))overset(+1)(P)overset(-2)(O_(2)^(-))` c) Locating atoms undergoing change in oxidation numbers `overset(0)(P_(4))rarroverset(-3)(P)H_(3)+H_(2)overset(+1)(P)O_(2)^(-)` d) Dividing the reaction into two two halves and balancing in acidic medium, separately Oxidation half - reaction : `P_(4)rarrH_(2)PO_(2)^(-)` Step 1 : Balance phosphorous atoms `P_(4)rarr4H_(2)PO_(2)^(-)` Step 2 : Balance oxygen atoms `P_(4)+8H_(2)Orarr4H_(2)PO_(2)^(-)` Step 3 : Balance hydrogen atoms `P_(4)+8H_(2)O+8OH^(-)rarr4H_(2)PO_(2)^(-)+8H_(2)O` Step 4 : Balance CHARGE `P_(4)+8OH^(-)rarr4H_(2)PO_(2)^(-)+4e^(-)".....(a)"` Reduction half - reaction : `P_(4)rarrPH_(3)` Step 1 : Balance phosphorous atoms `P_(4)rarr4PH_(3)` Stpe 2 : Balance hydrogen atoms `P_(4)+12H_(2)Orarr4PH_(3)+12OH^(-)` Step 3 : Balance charge `P_(4)+12H_(2)O+12E^(-)rarr4PH_(3)+12OH^(-)` e) Equalising the electrons and adding the two halves `"eq (a)"xx3+"eq (b)"xx"1, we get"` `{:(3P_(4)+24H_(2)Orarr12H_(2)PO_(2)^(-)+12e^(-)),(P_(4)+12H_(2)O+12e^(-)rarr4PH_(3)+12OH^(-)),(bar(4P_(4)+12OH^(-)+12H_(2)Orarr"")),(""4PH_(3)+12H_(2)PO_(2)^(-)),(I_(4)+3OH^(-)+3H_(2)OrarrPH_(3)+3H_(2)PO_(2)^(-)):}` This is the balanced equation. |
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